MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Since I haven't received a satisfactory answer to my initial question I'm going to ask a somewhat weaker one...

This time we say $X$ is a Vitali set in the closed interval $[0, 1]$ with respect to $\mathbb{Q}$ if $X$ is a selector of the partition of $[0, 1]$ canonically associated with the equivalence relation $x \in \mathbb{R}$ & $y \in \mathbb{R}$ & $x - y \in \mathbb{Q}$.

Recall a set $A$ has the property of Baire iff it can be represented in the form $F \triangle Q = (A \cup B) - (A \cap B)$, where $F$ is closed and $Q$ is of the first category (i.e. can be represented as a countable union of nowhere dense sets).

Let $r \in [0, 1]$, does there exist a Vitali set in the closed interval $[0, 1]$ with respect to $\mathbb{Q}$, $V$, such that $V \cup (V \oplus r)$ has the property of Baire where $V \oplus r$ = {$x + r : x \in V$ & $x + r \leq 1$} $\cup$
{$x + r - 1 : x \in V$ & $x + r > 1$}

The motivation for this is that no Vitali set has the property of Baire so an answer in the affirmative would answer my previous question (modified for the closed interval $[0, 1]$ and $\Gamma = \mathbb{Q}$) in the negative. However my guess is that there is no such Vitali set, but in which case I am interested in the proof which might begin by splitting the $V$ into a set of the first category and a null set and might say something about the spacing of elements of $V$.

One further question that would be of great help is if anyone knows any other properties of Vitali sets other than being non-measurable in the Lebesgue sense and not having the property of Baire.

share|cite|improve this question
up vote 3 down vote accepted

No, there does not exist such a $V$.

Let $W = V \cup (V \oplus r)$, and suppose $W = F \Delta Q$. Note that for $s \in {\mathbb Q}$, $W \cap (W \oplus s)$ is nonempty if and only if $s$ or $s-r$ or $s+r$ is an integer. But if $F$ contained an interval of positive length, $W \cap (W \oplus s)$ would be nonempty for all sufficiently small $|s|$. Thus $F$ is nowhere dense, and $W$ is of first category.
But this is impossible, because $[0,1] = \bigcup_{r \in {\mathbb Q}} (W \oplus r)$.

For your second question, the properties of Vitali sets (and finite unions of their translates) that this uses are:

1) there is a dense set of $r$ such that $V \cap (V \oplus r) = \emptyset$.

2) there is a countable set $S$ such that $\bigcup_{r \in S} (V \oplus r) = [0,1)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.